periodic angular acceleration of mechabism

hello

I am a simscape mulibody user

take a four bars mechanism, no external load

you apply a constant torque to the crank

I would expect a periodic angular acceleration but acceleration keeps growing

what's wrong ?

thanke and regards

Alberto Terzaghi

Answer 1

[David Goodmanson]

Hi Alberto,

(modified once more) If you apply a constant torque tau and advance the angle of the crank by an angle u, then you will increase the energy of the system by tau*u. I think we agree that as time evolves and there are many rotations, the increase in kintetic energy means that angular velocities keep going up without bound.

Consider a model that is simpler than the four bar model but which I think shares some important features. The model is basically a crankshaft and a piston. Bar 1 is of uniform density, mass M, length b. Bar 2 is massless and has length x0. One end of bar 1 has a pivot at the origin in the x-y plane; the two bars are attached with a pivot in the usual way; and the far end of bar 2 attaches with a pivot to a mass m over on the right, which is constrained to move along the x axis. There is a simple way way to constuct a mechanism to do this exactly, but let's say that bar 2 is so long compared to b that it can be considered to be horizontal, with negligible error. Conveniently, it's massless, so this works all right.

Let u be the angle between bar 1 and the x axis. The displacement of the mass is

x = x0 + b*cos(u)       so     x' = -b*sin(u)*u'

The Lagrangian for this system is

L = T-V = (1/2)*I*(u')^2 + (1/2)*m*(b*sin(u)*u')^2 + tau*u 

where tau is the constant torque, and I = (1/3)*M*b^2. Solving this gves

I*u'' + mb^2*( sin(u)*cos(u)*(u')^2 + sin(u)^2*u'' ) = tau

and rearranging this yields

u'' *(M/(3*m) + sin(u)^2)  =  tau/(mb^2) - sin(u)*cos(u)*(u')^2  

and it appears that as a function of time, u'' does increase without bound because (u')^2 on the right hand side does. I believe this is similar to a four bar system in that (u')^2 terms and the like happen in that system.

[Alberto Terzaghi]

Hi David

if you apply a constant torque the angular velocity keep going up without bound but the angular acceleration is constant if the moment of inertia is constant

in case of a four bar mechanism the moment of inertia, referred to the motor shaft ,is not constant but it is periodic thus also the instantaneous angular acceleration of the crank should be periodic

[David Goodmanson]

Hi Alberto,

I see what you mean, but there may be other factors involved, see modified answer above.

[Alberto Terzaghi]

hello David

I am not sure to follow your Euler Lagrange equations

suppose that the mass of the two bar 'a' and 'b' of your crankshaft mechanism are negligible, the mass of the piston is M

the moment of inertia referred to the motor shaft is

I = r^2*M = (a*cos(u) + b*sin(u))^2*M

I Is periodic so also the angular acceleration of the motor shaft A = T/I must be periodic

getting back to my four bar mechanism

I measured the angular position u = u(t) with the "transform measure" block

then I calculated the angular acceleration with two derivative blocks

I got the expected periodic acceleration

on the contrary if I click on the zalpha angular acceleration of the "trasnsform measure" block,

angular acceleration starts from zero and tenda to infinite

this is correct for kinetic energy but not for angular acceleration, in my opinion

cheer

Alberto

[Alberto Terzaghi]

modification, me too :)

of course the distance "r" of the piston from the center of rotation is a little bit more complicated but it is periodic

as far as I know every mechanism can be seen as a black box

on the otherside you can figure out the existance of a flywheel with variable but periodic moment of inertia

[David Goodmanson]

Hi Alberto,

I don't have simscape so I can not reprocuce any results you may have. I would say that the euler-lagrange equations don't lie. Consider the piston situation again as above. The kinetic energy of the bar, pivoted at its end, angle of rotation u, is

(1/2)*I*w^2 = (1/2)*((1/3)*M*b^2)*(u')^2

The velocity of the end of the bar is b*u', and there is a factor of sin(u) to get the correct velocity component for the of the piston of mass m. The kinetic energy of the piston is

(1/2)*m*v^2 = (1/2)*m*(b*sin(u)*u')^2

If you go with

K = (1/2)*Ieff*(u')^2,

the effective moment of inertia is

Ieff = (1/3)*M*b^2 + m*b^2*sin(u)^2

which is periodic in the angle as you said. However, it does not follow that for a constant torque the angular acceleration is periodic. As the bar rotates, the ampitude of the piston oscillation is always the same, namely b. But with constant torque, the angular velocity of the bar gets larger and larger. For each cycle, the piston oscillates at a not-quite-constant angular velocity w, but that w is increasing all the time. The acceleration of the piston is approximately w^2*b, which increases without bound. I believe similar behavior will happen with a four bar linkage at constant torque.